Suppose I have a finite field $F_{p^d}$, and I have a polynomial $f$ which is of degree $n$ and irreducible. I have a feeling that the splitting field of $f$ over $F_{p^d}$ is $F_{p^{dn}}$, but I am not sure how to show this.

I think that since finite fields are perfect, every irreducible polynomial is separable, meaning that the splitting field of $f$ is $F_{p^d}(\alpha)$ where the degree of the minimal polynomial of $\alpha$ is $n$, and so the splitting field is like a degree $n$ vector space over $F_{p^d}$.

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    $\begingroup$ For all $m$ and for all agebraically closed fields $k$ of characteristic $p$ there is exactly one subfield $F\subseteq k$ such that $\lvert F\rvert=p^m$. Namely, it can be characterized as the splitting field of the polynomial $x^{p^m}-x$ over the base field. For degree reasons, every root $\alpha$ of $f$ must satisfy $F_{p^d}(\alpha)=F_{p^{dn}}$. $\endgroup$ – Saucy O'Path May 8 at 0:07

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